1007. Maximum Subsequence Sum (25)

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Given a sequence of K integers { N1, N2, …, NK }. A continuous subsequence is defined to be { Ni, Ni+1, …, Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

思路：题意是求序列中最大子序列和，并要标记第一个最大子序列的开始元素和结束元素，遍历序列，若当前子序列和大于最大值，就更新开始元素和结束元素和最大值，若当前子序列小于零就将子序列和置零并将记录临时开始元素为下一个，算法复杂度为O（n）。

Code：

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#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,i,arr[10010];    
    cin>>n;
    int flag=0,st=1,ed=n,max=-1,sum=0,tpst=1;
    for(i=1;i<=n;++i)
    {
        cin>>arr[i];
        if(arr[i]>=0)
            flag=1;
    }
    if(flag==0)
        cout<<0<<" "<<arr[st]<<" "<<arr[ed]<<endl;
    else
    {
        for(i=1;i<=n;++i)
        {
            sum+=arr[i];
            if(sum>max)
            {
                max=sum;
                st=tpst;
                ed=i;
            }
            if(sum<0)
            {
                sum=0;
                tpst=i+1;
            }            
        }
        cout<<max<<" "<<arr[st]<<" "<<arr[ed]<<endl;
    }
    return 0;
}

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